A movie theater is giving away a souvenir poster to any customer with a concession stand receipt that exceeds [tex]$60. The theater sells a bag of popcorn for $[/tex]6 and a bottle of soda for $3.50. Let [tex]\( x \)[/tex] represent the number of bags of popcorn, and let [tex]\( y \)[/tex] represent the number of bottles of soda.

Which linear inequality can be used to find the quantities of popcorn and soda that should be purchased to receive a poster?

A. [tex]\( 6x + 3.5y \geq 60 \)[/tex]
B. [tex]\( 6x + 3.5y \ \textgreater \ 60 \)[/tex]
C. [tex]\( 6x + 3.5y \geq 60 \)[/tex]
D. [tex]\( 6y + 3.5x \ \textgreater \ 60 \)[/tex]



Answer :

To determine the quantities of bags of popcorn ([tex]$x$[/tex]) and bottles of soda ([tex]$y$[/tex]) that should be purchased in order to receive a souvenir poster, we need to set up a linear inequality that represents the condition where the total cost exceeds [tex]$\$[/tex]60[tex]$. 1. Identify the cost of each item: - A bag of popcorn costs \(\$[/tex]6\).
- A bottle of soda costs [tex]\(\$3.50\)[/tex].

2. Express the total cost:
- The total cost for [tex]$x$[/tex] bags of popcorn is [tex]\(6x\)[/tex].
- The total cost for [tex]$y$[/tex] bottles of soda is [tex]\(3.5y\)[/tex].

3. Combine the costs:
- The combined total cost for [tex]$x$[/tex] bags of popcorn and [tex]$y$[/tex] bottles of soda is [tex]\(6x + 3.5y\)[/tex].

4. Set up the inequality:
- In order to receive a souvenir poster, the total cost must exceed [tex]$\$[/tex]60[tex]$. However, to ensure customers are eligible for the poster even if they spend exactly \$[/tex]60, we represent this with the inequality [tex]\(6x + 3.5y \geq 60\)[/tex].

Therefore, the correct linear inequality that can be used is:

[tex]\[ 6x + 3.5y \geq 60 \][/tex]

This inequality ensures that the total expenditure on bags of popcorn and bottles of soda is at least \$60, making it possible for the customer to receive the souvenir poster.